PASSAGE:

Let f be a function defined on the set of whole no. and takes non negative integer values such that , f(0) = 0,f(1) = 1,f(n) = f(n-1) + f(n-2), for n>=2

(1)               We must have

(a)    f(n+1)>=f(n) for all n>=2  (b) f(n+1)<=f(n) for all n>=2

(c) f(n+1)>f(n) for all n>=2    (d) f(n+1)<f(n) for all n>=2

(2)               The no. Of non negative integer n, such that f(f(n)) = f(n),is

(a) 3       (b) 4  (c) 5  (d) 6

(3)               f(n+2) is equal to

(a)   

(b)   1 +

(c)   

(d)   None of these.

Ans = (1) c

           (2) b

          (3) b

isnt the series..the fibonacci numbers...??

1. c

2. a

3. Question unclear?

yes, it is indeed the fibonnaci series. Every term is the sum of the previous two.

The series is:

0,1,1,2,3,5,8,13,21,34,55..........

from the series it is clear that for n>=2, the series increases, so for qn 1 (c) is right.

for q2

f(f(n)) = f(n)

=>f({f(n)}-1) + f({f(n)} - 2) = f(n)

if f(n)=t

=> f(t-1) + f(t-2) = t

=> f(t) = t...

which is true only for t = 0,1 and 5 so ans is (a) :)